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Question: How do you convert \(3\sqrt 2 - 3\sqrt 2 i\) to polar form?...

How do you convert 32−32i3\sqrt 2 - 3\sqrt 2 i to polar form?

Explanation

Solution

We have been given a complex number which is to be converted into polar form. A complex number can be represented in various forms such as rectangular, exponential and polar forms.
Rectangular form: a+bia + bi
Exponential form: reiθr{e^{i\theta }}
Polar form: r(cos⁡θ+isin⁡θ)r(\cos \theta + i\sin \theta )
We can see that we have to convert the rectangular form of the complex number to the polar form.

Complete step by step solution:
We have been given a complex number 32−32i3\sqrt 2 - 3\sqrt 2 i in rectangular form. We have to convert this complex number into polar form without changing the value.
The polar form of a complex number is given as r(cos⁡θ+isin⁡θ)r(\cos \theta + i\sin \theta ).
When we plot a complex number in the complex plane, θ\theta is the angle between the complex vector and the real axis, known as the argument of the complex number. For a complex number a+bia + bi, θ\theta is given as tan⁡θ=ba\tan \theta = \dfrac{b}{a}.
On comparing the given complex number with general form, we get,
a=32a = 3\sqrt 2 and b=−32b = - 3\sqrt 2 .
Thus, tan⁡θ=ba=−3232=−1⇒θ=tan⁡−1(−1)=−π4\tan \theta = \dfrac{b}{a} = \dfrac{{ - 3\sqrt 2 }}{{3\sqrt 2 }} = - 1 \Rightarrow \theta = {\tan ^{ - 1}}\left( { - 1} \right) = - \dfrac{\pi }{4}
The value of rr is the absolute value of the given complex number given as, r=a2+b2r = \sqrt {{a^2} + {b^2}}
Thus, r=(32)2+(−32)2=18+18=36=6r = \sqrt {{{\left( {3\sqrt 2 } \right)}^2} + {{\left( { - 3\sqrt 2 } \right)}^2}} = \sqrt {18 + 18} = \sqrt {36} = 6
We can put the obtained value of rr and θ\theta to write the complex number as,
r(cos⁡θ+isin⁡θ)=6(cos⁡(−π4)+isin⁡(−π4))r(\cos \theta + i\sin \theta ) = 6\left( {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right)

Hence, in polar form 32−32i=6(cos⁡(−π4)+isin⁡(−π4))3\sqrt 2 - 3\sqrt 2 i = 6\left( {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right)

Note: For calculation of the angle θ\theta we found the inverse tan of the ratio of the imaginary part and the real part of the complex number. The angle is taken to be acute, it can be negative or zero. The value of rr is the modulus or the absolute value of the given complex number. If we simplify the polar form by solving the sine and cosine ratios of the angle, we will arrive back to the rectangular form.